Question

Milk enters at 25°C through the inner pipe of a concentric double pipe heat exchanger. Hot water enters at 82.5°C and flows countercurrently (flow rate = 1.2 kg/s) through the annular region. The diameter of the inner pipe, length of the pipe, and average overall heat transfer coefficient are 60 mm, 6 m, and 2100 W.m\(^{-2}\).K\(^{-1}\), respectively. The average values of specific heat capacity of water and milk are 4.18 kJ.kg\(^{-1}\).K\(^{-1}\) and 3.95 kJ.kg\(^{-1}\).K\(^{-1}\), respectively. The effectiveness of the heat exchanger is 0.572 and the NTU is 1.01. Assuming the steady-state condition and considering \( \pi \) as 3.14, the temperature of hot water at the exit of the pipe is _________ °C. (Rounded off to 1 decimal place)

Hint:

To calculate the outlet temperature of hot water, use the effectiveness of the heat exchanger and the maximum possible heat transfer.

Answer 1

We are given the following parameters:
Temperature of hot water entering (\( T_{{h,in}} \)) = 82.5°C
Flow rate of hot water (\( \dot{m}_h \)) = 1.2 kg/s
Overall heat transfer coefficient (\( U \)) = 2100 W.m\(^{-2}\).K\(^{-1}\)
Length of heat exchanger (\( L \)) = 6 m
Diameter of inner pipe (\( D_i \)) = 60 mm = 0.06 m
Effectiveness (\( \epsilon \)) = 0.572
NTU = 1.01
Specific heat capacity of water (\( C_{p,h} \)) = 4.18 kJ/kg.K = 4180 J/kg.K
Specific heat capacity of milk (\( C_{p,c} \)) = 3.95 kJ/kg.K = 3950 J/kg.K
Inlet temperature of milk (\( T_{{c,in}} \)) = 25°C
Step 1: Calculate the heat exchanged using effectiveness
The formula for the heat exchanged in a heat exchanger is given by: \[ Q = \epsilon \times Q_{{max}} \] Where \( Q_{{max}} \) is the maximum possible heat transfer and is calculated as: \[ Q_{{max}} = \dot{m}_h C_{p,h} (T_{{h,in}} - T_{{c,in}}) \] Substitute the known values: \[ Q_{{max}} = 1.2 \times 4180 \times (82.5 - 25) = 1.2 \times 4180 \times 57.5 = 287220 \, {J/s} = 287.22 \, {kW} \] Now, calculate the heat exchanged using the effectiveness: \[ Q = 0.572 \times 287.22 = 164.41 \, {kW} \] Step 2: Use the heat exchanged to calculate the outlet temperature of hot water The heat exchanged is also given by: \[ Q = \dot{m}_h C_{p,h} (T_{{h,in}} - T_{{h,out}}) \] Where \( T_{{h,out}} \) is the outlet temperature of the hot water. Rearrange to solve for \( T_{{h,out}} \): \[ T_{{h,out}} = T_{{h,in}} - \frac{Q}{\dot{m}_h C_{p,h}} \] Substitute the known values: \[ T_{{h,out}} = 82.5 - \frac{164.41 \times 1000}{1.2 \times 4180} = 82.5 - \frac{164410}{5016} = 82.5 - 15.0 = 67.5^\circ {C} \] Thus, the temperature of hot water at the exit of the pipe is 67.5°C.